Thank you for reporting, we will resolve it shortly
Q.
The equivalent capacitance between $A$ and $B$ in the circuit given below, is :
JEE MainJEE Main 2018Electrostatic Potential and Capacitance
Solution:
The given circuit is as shown below:
Here capacitance $C_{2}, C_{3}$ and $C_{4}$ are connected in parallel. Therefore,
$C'=C_{2}+C_{3}+C_{4}=5 \mu F +5 \mu F +2 \mu F =12 \mu F$
Also, capacitance $C_{5}$ and $C_{6}$ are in parallel. Therefore,
$C''=C_{5}+C_{6}=2 \mu F +4 \mu F =6 \mu F$
Therefore, equivalent capacitance of the circuit is
$\frac{1}{C_{e q}}=\frac{1}{C_{1}}+\frac{1}{C'}+\frac{1}{C''}=\frac{1}{6 \mu F }+\frac{1}{12 \mu F }+\frac{1}{6 \mu F }$
$=\frac{2+1+2}{12 \mu F }=\frac{5}{12 \mu F }$
$\Rightarrow C_{\text {eq }}=\frac{12}{5} \mu F =2.4\, \mu F$
